Engr Help

Center of Curvature (Locating center of curvature of osculating circle)

Example: A particle, P, is moving along a path in the x-y plane as shown in the figure below.

Its position vector, r_p , given at any time t is:

r_p(t) = ˗ 3t i + (t² ˗ 3t ˗2) j ft

Find the location of the center of curvature , r_c at t = 1 sec.

Solution: Use r_c = r_p + ρ e_n r_p(1) = ˗ 3i ˗ 4 j

Strategy: 1. First find e_n using e_n = de_t/dt then find ρ using a_n = v² / ρ

Here e_t is the unit tangential vector, e_n is the unit normal vector

ρ is the radius of curvature, and v is the speed of the particle

dr_p(t)/dt = ˗3 i + (2t ˗ 3) j So | dr_p(t)/dt | = √ (18 + 4t² ˗12t )

So e_t = ˗3 i + (2t ˗ 3) j / √ (18 + 4t² ˗12t )

de_t / dt = 2 j / √ (18 + 4t² ˗12t ) ˗ (1/2) [˗3 i + (2t ˗ 3) j ] [ 8t ˗ 12 ] / (18 + 4t² ˗12t )^3/2

de_t / dt = { [ 12t ˗ 18] i + [ 8 t² ˗ 24t + 36 ˗ 2(2t ˗ 3)²] j } / (18 + 4t² ˗12t )^3/2

At t = 1 sec de_t / dt = ˗ 6 i + 18 j / (10) ^3/2 = ˗ 6 i + 18 j / (10) √ 10 = (3/5) [ ˗ i + 3 j ] /√10

And | de_t / dt | = 3/5 After algebra e_n= de_t / dt / | de_t / dt | = [ ˗ i + 3 j ] /√10

Use a_n = v² / ρ or ρ = v² / a_n At t = 1 sec v = | dr_p/dt | = √ 10